Advanced Function Transformations

Derive the complete point map

Let

Suppose $(u,v)$ lies on the graph of $f$, so $v=f(u)$. To use the same original input $u$ inside $g$, solve $B(x-h)=u$. Then

The complete point mapping is therefore

Horizontal distances scale by $1/|B|$, with a reversal when $B<0$. Vertical distances scale by $|A|$, with a reversal when $A<0$. The parameters $h,k$ locate the translations in this factored form.

If $A=0$, the output is constant on the original composite’s domain, not necessarily on every real input. If $B=0$, the inner expression is constant and must belong to the domain of $f$. The ordinary invertible point map does not apply to those degenerate cases.

Map a radical graph with two reversals

Take $f(u)=\sqrt u$ and

Here $A=-2$, $B=-3$, $h=1$, $k=4$, giving

The parent points $(0,0),(1,1),(4,2),(9,3)$ become

Each image point can be checked in the displayed formula. The domain comes from $-3(x-1)\ge0$, so $x\le1$. Since the square root is nonnegative, $g(x)\le4$. Every value at or below 4 is attained: for a desired output $y\le4$, set $\sqrt{-3(x-1)}=(4-y)/2$ and solve for $x$.

The graph’s endpoint is $(1,4)$, not $(3,4)$ or $(1,-4)$. Working with the point map prevents the two negative factors from being confused.

Transform intervals with their order and endpoints

Suppose a parent function is restricted to domain $[-3,6]$ and has range $[0,5]$. Under the same mapping,

The input endpoints $u=-3,6$ map to $x=2,-1$. Reorder them when writing the interval: the transformed domain is $[-1,2]$. The output endpoints 0 and 5 map to 4 and −6, so the range is $[-6,4]$.

Negative scale factors reverse endpoint order but do not change whether a given endpoint is included. If the original domain were $(-3,6]$, its image would be $[-1,2)$: the image of the excluded endpoint remains excluded.

This example concerns a specified bounded parent domain and range. It is a different parent-function situation from the unrestricted square-root example above. Domain and range information must accompany the chosen function, not be borrowed from a neighbouring example.

Factor the inner expression before reading a shift

For

factor the input as $3(x+2)$. In the standard form, $B=3$ and $h=-2$, giving the map

The horizontal translation is −2, not +6. One valid construction compresses the parent graph horizontally by a factor of $1/3$, then shifts left two units. Shifting first and then compressing by the same stated amounts would usually give a different function.

The algebra determines the correct mapping. A verbal list of transformations is useful only when its order produces that mapping. OpenStax’s transformation chapter provides the basic transformations being combined here.

Distinguish folding outputs from folding inputs

For $g(x)=|f(x)|$, keep each input and replace its output by its absolute value. Portions below the x-axis reflect upward; portions on or above it stay unchanged. Thus

By contrast, $g(x)=f(|x|)$ uses the nonnegative side of the parent function as the source and mirrors it across the y-axis. For $f(x)=x-1$,

whereas $|f(x)|=|x-1|$. Their vertices are $(0,-1)$ and $(1,0)$ respectively, so they are plainly different graphs.

The domain of $f(|x|)$ is $\{x:|x|\in\operatorname{dom}f\}$. If the parent exists only for negative inputs, folding the input might leave no allowed input. Do not assume every absolute-value transformation automatically has domain $\mathbb R$.

Map asymptotes and preserve forbidden inputs

Start with $f(u)=1/u$ and define

The input $u=0$ maps to $x=2$, so the vertical asymptote is $x=2$. The horizontal asymptote $v=0$ maps to $y=1$. Directly,

at both ends. The point map is $(u,v)\mapsto(2+u/2,1+3v)$.

If the parent had an additional excluded point, that exclusion would map too. A simplified transformed formula should not silently restore it. Asymptotes describe limiting behaviour; holes describe missing points with finite nearby limits.

Check logarithms and exponentials after transformation

The transformed logarithm $\ln(4-2x)$ requires $4-2x>0$, giving $x<2$. Its vertical boundary is excluded. This strict condition differs from the inclusive endpoint of a square root, whose argument may equal zero.

For

the domain is all real numbers and the range is $(-\infty,5)$. As $x$ increases, $3^{-(x+1)}$ decreases, and multiplication by −2 reverses that trend. The final function is increasing, although both a horizontal and vertical reflection occur in its construction.

The output approaches 5 from below as $x\to+\infty$. Setting $q(x)=0$ gives $3^{-(x+1)}=5/2$, hence $x=-1-\ln(5/2)/\ln3$. A graph label should be placed using this actual intercept, not an approximate visual guess.

Related topics

Composite and Inverse Functions: Advanced Problems reverses and combines these mappings while retaining the correct domains. Asymptotes develops the limiting behaviour of transformed rational functions.